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Unformatted text preview: xy -plane is x2 + y 2 = 1, √ 1−x2 1 4−3y 2 V =4 dz dy dx 4x2 +y 2 0 0 20. The projection of the curve of intersection onto the xy -plane is 2x2 + y 2 = 4, √ 2 √ 4−2x2 8−x2 −y 2 V =4 dz dy dx 0 0 3x2 +y 2 √ 9−x2 /3 3 x+3 dz dy dx 21. V = 2 −3 0 0 √ 1−x2 1 √ 1−x2 22. V = 8 dz dy dx 0 0 0 Exercise Set 16.5 588 z 23. (a) z (b) (0, 9, 9) (0, 0, 1) y x (1, 0, 0) (3, 9, 0) y x z (c) (0, 0, 1) y (1, 2, 0) x z 24. (a) z (b) (0, 0, 2) (0, 0, 2) (0, 2, 0) y y (3, 9, 0) (2, 0, 0) x x z (c) (0, 0, 4) y (2, 2, 0) x 1−x 1 1−x−y 25. V = 1−x 1 1−x−y dz dy dx = 1/6, fave = 6 0 0 0 (x + y + z ) dz dy dx = 0 0 0 26. The integrand is an odd function of each of x, y , and z , so the answer is zero. 3 4 589 Chapter 16 b(1−x/a) a c(1−x/a−y/b) a(1−y/b) b 27. (a) c(1−x/a−y/b) dz dy dx, 0 0 a(1−z/c) c dz dx dy , 0 0 b(1−x/a−z/c) 0 0 c(1−x/a) a b(1−x/a−z/c) dy dx dz , 0 0 b(1−z/c) c dy dz dx, 0 0 a(1−y/b−z/c) 0 0 c(1−y/b) b a(1−y/b−z/c) dx dy dz , 0 0 dx dz dy 0 0 0 0 (b) Use the ﬁrst integral in part (a) to get b(1−x/a) a 0 xy − dy dx = a b c 1− 0 √ √ 9−x2 3 0 2 x 1 bc 1 − 2 a dx = 1 abc 6 9−x2 −y 2 f (x, y, z )dz dy dx 28. (a) 0 0 0 x/2 4 2 0 0 b √ √ c x2 0 1−x2 /a2 −y 2 /b2 29. V = 8 dz dy dx 0 0 b f (x, y, z )dz dy dx 0 1−x2 /a2 4−y (c) 0 a 4−x2 2 f (x, y, z )dz dy dx (b) 0 d b d f (x)g (y )h(z )dz dy dx = 30. a a c k f (x)g (y ) a h(z )dz dy dx c k b d f (x) = a g (y )dy dx c b a 1 −1 y 2 dy 0 sin z dz = (0)(1/3)(1) = 0 ln 3 e2x dx 0 h(z )dz k 0 1 (b) g (y )dy c π /2 1 x dx 31. (a) d f (x)dx = h(z )dz k ln 2 ey dy 0 e−z dz = [(e2 − 1)/2](2)(1/2) = (e2 − 1)/2 0 32. (a) At any point outside the closed sphere {x2 + y 2 + z 2 ≤ 1} the integrand is negative, so to maximize the integral it suﬃces to include all points inside the sphere; hence the maximum value is taken on the region G = {x2 + y 2 + z 2 ≤ 1}. (b) 8π 15 EXERCISE SET 16.6 1. Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum is 5(0 − a) + 10(5 − a) + 20(10 − a) = 0 for equilibrium, so 250 − 35a = 0, a = 50/7. The fulcrum should be placed 50/7 ft to the right of m1 . Exercise Set 16.6 590 2. At equilibrium, 10(0 − 4) + 3(2 − 4) + 4(3 − 4) + m(6 − 4) = 0, m = 25 1 1 0 0 1+x 0 1 2 4. A = 2, x = ¯ 1 y dy dx = 0 0 −1 −1−x x dy dx = 0, similarly y = 0. ¯ −1+x 0 x x dA = 0 x 1 x dy dx = 1/3, R 1 2 1−x 1 x dy dx + 1 5. A = 1/2, 1 1 , y= ¯ 2 x dy dx = 3. A = 1, x = ¯ y dA = 0 y dy dx = 1/6; 0 R 0 centroid (2/3, 1/3) x2 1 0 0 x2 1 dy dx = 1/3, 6. A = x dA = x dy dx = 1/4, 0 0 R x2 1 y dA = y dy dx = 1/10; centroid (3/4, 3/10) 0 0 R 2−x2 1 dy dx = 7/6, 7. A = x 0 x dA = x dy dx = 5/12, x 0 R 2−x2 1 y dA = y dy dx = 19/15; centroid (5/14, 38/35) x 0 R √ 1−x2 1 π 8. A = , 4 2−x2 1 x dA = x dy dx = 0 0 R 4 4 1 ,x= ¯ , y= ¯ by symmetry 3 3π 3π 9. x = 0 from the symmetry of the region, ¯ A= π 1 π (b2 − a2 ), 2 b r2 sin θ dr dθ = y dA = 0 R a 23 4(b3 − a3 ) (b − a3 ); centroid x = 0, y = . ¯ ¯ 3 3π (b2 − a2 ) 10. y = 0 from the symmetry of the region, A = πa2 /2, ¯ π /2 a r2 cos θ dr dθ = 2a3 /3; centroid x dA = −π/2 R 0 √ x 1 11. M = √ x 1 (x + y )dy dx = 13/20, Mx = 0 0 My = (x + y )y dy dx = 3/10, 0 √ x 1 4a ,0 3π 0 (x + y )x dy dx = 19/42, x = My /M = 190/273, y = Mx /M = 6/13; ¯ ¯ 0 0 the mass is 13/20 and the center of gravity is at (190/273, 6/13). π sin x y dy dx = π/4, x = π/2 from the symmetry of the density and the region, ¯ 12. M = 0 0 π sin x y 2 dy dx = 4/9, y = Mx /M = ¯ Mx = 0 0 16 ; mass π/4, center of gravity 9π π 16 , . 2 9π 591 Chapter 16 π /2 a r3 sin θ cos θ dr dθ = a4 /8, x = y from the symmetry of the density and the ¯ ¯ 13. M = 0 0 π /2 a r4 sin θ cos2 θ dr dθ = a5 /15, x = 8a/15; mass a4 /8, center of gravity ¯ region, My = 0 0 (8a/15, 8a/15). π 1 r3 dr dθ = π/4, x = 0 from the symmetry of density and region, ¯ 14. M = 0 0 π 1 r4 sin θ dr dθ = 2/5, y = ¯ Mx = 0 0 1 1 1 x dz dy dx = 15. V = 1, x = ¯ 0 0 0 8 ; mass π/4, center of gravity 5π 1 1 , similarly y = z = ; centroid ¯¯ 2 2 1 z dz dy dx = rz dr dθ dz = 2π , 0 G centroid = (0, 0, 1) 8 . 5π 111 ,, 222 2π 2 16. V = πr2 h = 2π , x = y = 0 by symmetry, ¯¯ 0, 0 0 17. x = y = z from the symmetry of the region, V = 1/6, ¯¯¯ x= ¯ 1−x 1 1 V 1−x−y x dz dy dx = (6)(1/24) = 1/4; centroid (1/4, 1/4, 1/4) 0 0 0 18. The solid is described by −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y 2 , 0 ≤ x ≤ 1 − z ; 1−y 2 1 1−z V= dx dz dy = −1 z= ¯ 0 0 1−y 2 1 1 V 1 4 ,x = ¯ 5 V 1−z 0 0 1−z x dx dz dy = −1 0 2 ; the centroid is 7 z dx dz dy = −1 1−y 2 1 0 5 , y = 0 by symmetry, ¯ 14 2 5 , 0, . 14 7 19. x = 1/2 and y = 0 from the symmetry of the region, ¯ ¯ 1 1 1 V= dz dy dx = 4/3, z = ¯ −1 0 y2 1 V z dV = (3/4)(4/5) = 3/5; centroid (1/2, 0, 3/5) G 20. x = y from the symmetry of the region, ¯¯ 2 xy 2 dz dy dx = 4, x = ¯ V= 0 z= ¯ 0 1 V 0 1 V x dV = (1/4)(16/...
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